Timeline for Embedding a cancellative monoid into another in such a way that $|X-x|=|X|$, where $X$ is a fixed finite set and $x\in X$
Current License: CC BY-SA 3.0
7 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
Mar 14, 2016 at 22:18 | comment | added | Salvo Tringali | Sorry for the delay, I was in a hurry this afternoon and didn't realize that there were at least three things I had to fix in the old formulation. In particular, I would/should have asked if $\mathbb B$ can be embedded into $\mathbb A$ in such a way that at least one $x\in X$ becomes right-subtractive. @BenjaminSteinberg. Yes, and if $\mathbb A$ is a canc. monoid, then the right-subtractive elements of a finite set $X$ that contains the identity of $\mathbb A$ are precisely the left-invertible elements of $\mathbb A$ (hence $X^{\rm rs}=\mathbb A^\times$, from what you made me note yesterday). | |
Mar 14, 2016 at 21:59 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Changed a word in the title
|
Mar 14, 2016 at 21:49 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
Fixed a couple of issues in the formulation of the question
|
Mar 14, 2016 at 17:47 | comment | added | Benjamin Steinberg | If $X$ is just the identity, then isn't $X$-subtractive the same as left-invertible? | |
Mar 14, 2016 at 14:40 | history | edited | Salvo Tringali | CC BY-SA 3.0 |
added 10 characters in body
|
Mar 14, 2016 at 14:34 | history | asked | Salvo Tringali | CC BY-SA 3.0 |